$AB=a$ and $AC=b$ are the sides of $\triangle ABC$. $P$ is a point on $AB$ and $Q$ is a point on $BC$ such that $\frac{AP}{PB}=\frac{1}{2}$ and $\frac{BQ}{QC}=\frac{1}{2}$. If the point of intersection of $AQ$ and $CP$ is $D$ and the area of $\triangle BCD$ is $7$ square units,then the area of the $\triangle ABC$ (in the same square units) is

In a quadrilateral $ABCD$,if $P$ and $Q$ are the midpoints of $\overline{BC}$ and $\overline{AD}$ respectively,then $\vec{AB} + \vec{DC} = \dots$

If two vectors $\vec{a}$ and $\vec{b}$ which are perpendicular to each other are such that $|\vec{a}|=8$ and $|\vec{b}|=3$, then $|\vec{a}-2\vec{b}|=$

If $\vec{a}, \vec{b}, \vec{c}$ are vectors such that $\vec{a}+\vec{b}+\vec{c}=\vec{0}$ and $|\vec{a}|=7, |\vec{b}|=5, |\vec{c}|=3$,then the angle between vector $\vec{b}$ and $\vec{c}$ is: (in $^{\circ}$)

Let a unit vector $\hat{u}=x \hat{i}+y \hat{j}+z \hat{k}$ make angles $\frac{\pi}{2}, \frac{\pi}{3}$ and $\frac{2 \pi}{3}$ with the vectors $\frac{1}{\sqrt{2}} \hat{i}+\frac{1}{\sqrt{2}} \hat{k}, \frac{1}{\sqrt{2}} \hat{j}+\frac{1}{\sqrt{2}} \hat{k}$ and $\frac{1}{\sqrt{2}} \hat{i}+\frac{1}{\sqrt{2}} \hat{j}$ respectively. If $\overrightarrow{v}=\frac{1}{\sqrt{2}} \hat{i}+\frac{1}{\sqrt{2}} \hat{j}+\frac{1}{\sqrt{2}} \hat{k}$,then $|\hat{u}-\overrightarrow{v}|^2$ is equal to

Let $\vec{a}=\hat{i}+2\hat{j}+\hat{k}$ and $\vec{b}=2\hat{i}+\hat{j}-\hat{k}$. Let $\hat{c}$ be a unit vector in the plane of the vectors $\vec{a}$ and $\vec{b}$ and be perpendicular to $\vec{a}$. Then such a vector $\hat{c}$ is :